6688 Statistics S6 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

June 2005 6688 Statistics S6

Mark Scheme Question Number

Scheme

Marks

1. 2. 3.

H0 : Medianm – Medians = 0; H1 : Medianm – Medians > 0 8 + & 2 – N=10 P( 2 p = 0.5, n =10) = 0.0547 > 0.05 Insufficient evidence to reject H0. Students do not carry out calculations more accurately when listening to music. H0 : Median = 20; H1 : Median 20

x - median +10 -8 -5 +1 +7 -3 -4 -6 +2 Rank 9 8 5 1 7 3 4 6 2

S+ = 19 ( S- = 26) n = 9 CR : S 5 Since 19 is not in the critical region there is insufficient evidence to reject H0. The claim is justified on this evidence. H0 : Median time of girls = Median time of boys H1 : Median time of girls Median time of boys both n1 = 25, n2 = 25

2656.25 637.5,N12

125252525

2

1252525N

,T

291252656

5637704.

.

.Z

Accept 704 0.5

Since 1.29 is not in the critical region (z > 1.96) there is no evidence to suggest that boys are quicker at French translation than girls.

B1 M1 A1 M1 A1 A1

(6)

B1 M1 M1 A1 A1 B1 A1

(7)

B1 M1 A1 A1 M1 A1 A1

(7)

6688 Statistics S6 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

graph 2 scales & limits

4. 5. (a)

(b) (c) (d) (e)

C.F. =

24

41191

24

6553504613431348 22 .....

SST = 1543.9043 -

32981724

41191 2

..

SSA = 63659

24

411916553504613431348

6

1 22222 .

.....

Source df

s.s mss Ratio

Between areas 3 9.6365 3.2122 8.35 Residual 20 7.6933 0.3847

Total 23 17.3298 H0 : Not all means are equal (Assume = 0.05) F3, 20 = 3.10 (4.94 for 1%) Since 8.35 is in the critical region there is evidence that there is a difference in the mean yields between areas.

06405010

32 φ .ˆ

UWL = 0.064 + 1.96 59

93600640 .. = 0.1318….

UAL = 0.064 + 2.5758 50

93600640 .. = 0.153156…

Graph (Limits and scales) Target value is zero; Company not concerned if p tends to zero. Graph (Points) All points below warning limit so production is in control.

M1 A1 M1 A1 B1 B1 M1 A1 B1 B1 A1 (11) B1 (1) M1 B1 A1 B1 A1 (5) B2 (2) B1 B1 (2) B2 (2) B1 (1)

df Residual

Ratio

6688 Statistics S6 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

0

0.05

0.1

0.15

0.2

0 1 2 3 4 5 6 7 8 9 10

Sample number

Fra

ctio

n de

fect

ive

Scales & labels B1, Limits B1 2 Points B2 2

Warning limit

Action limit

6688 Statistics S6 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 (a) (b)

Randomised Block Design

SSO = 12

886287297302

4

1 2222 = 29.17

SSM = 12

886186271250179

3

1 22222 = 2109.67

Source df s.s. MSS Ratio

Operators 2 29.17 14.59 0.75 Machines 3 2109.67 703.22 36.12 Residual 6 116.83 19.47

Total 11 2255.67 (i) H0 : A = B = C = D ; H1 : Not all means are equal = 0.05 (say) (0.05) = 4.76 ( (0.01) = 9.78) 3

6F 36F

Since 36.12 is in the critical region there is evidence of differences between machines (ii) H0 : 1 = 2 = 3 ; H1 : Not all means are equal = 0.05 (say) (0.05) = 5.14 ( (0.01) = 10.90) 2

6F 26F

Since 0.75 is not in the critical region there is insufficient evidence to reject H0. There are no differences in the mean quality of the operators.

B1 (1) B1 B1 B1 B1 M1 A1 A1 B1 B1 B1 A1 B1 B1 A1 (14)

df Residual Ratios

6688 Statistics S6 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics

Question Number

Scheme

Marks

7. (a) (b) (c) (d)

Sxy = 6493576 - 20

929714061= -42679.85

= 9513998

8542679

.

. = -3.048789… -3.05

= 20

140610487893

20

9297 .... = 2608.303167… 2608.30

y = 2608.30 – 3.05x

RSS = 164592.55 - 9513998

)8542679( 2

.

. = 34470.67686

95% CI is given by

- 3.048789… 2.101 951399818

6768634470

.

.

i.e. –3.83 & -2.27 -3.00 is in the CI Assumption is justified. N(0, 2) Plot residuals (y - y ) against x Residuals randomly scattered about x axis model justified

B1 M1 A1 M1 A1 (5) M1 A1 M1 B1 A1

A1 (6) B1 B1 (2) B1 B1 B1 (3)